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Problems · Problem 7.2

Q.Justify that the reaction: 2Na(s) + H2(g) → 2NaH(s) is a redox change.

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Sodium metal loses electrons (oxidation) while hydrogen gas gains electrons (reduction) to form sodium hydride, where hydrogen exists as the hydride ion H−\text{H}^-. This electron transfer confirms the reaction is redox.

The heart of recognizing a redox reaction lies in tracking electron movement. Redox is shorthand for reduction-oxidation: one species must lose electrons (oxidation) while another gains them (reduction). These two half-processes always occur together because electrons don't vanish—they simply relocate.

In this reaction, we need to assign oxidation states to every element on both sides and watch for changes.

Assigning oxidation states

  1. Elemental forms have zero oxidation state.

    Both Na(s)\text{Na}(s) and H2(g)\text{H}_2(g) are elements in their standard states, so:

    • Sodium: Na=0\text{Na} = 0
    • Hydrogen: H=0\text{H} = 0
  2. In sodium hydride NaH\text{NaH}, we apply the rules for ionic compounds.

    Sodium is a Group 1 metal and always takes a +1+1 oxidation state in compounds. For the compound to be electrically neutral:

Na+1+Hx=0  ⟹  x=−1\text{Na}^{+1} + \text{H}^{x} = 0 \implies x = -1

So hydrogen exists as the hydride ion H−\text{H}^- with oxidation state −1-1.

Note

This is unusual! Hydrogen typically has oxidation state +1+1 in most compounds (like H2O\text{H}_2\text{O} or HCl\text{HCl}). Only when bonded to metals does it become H−\text{H}^-.

Identifying oxidation and reduction

  1. Track the changes in oxidation state:

    SpeciesInitial stateFinal stateChange
    Na\text{Na}00+1+1Loses 1 electron (oxidised)
    H\text{H}00−1-1Gains 1 electron (reduced)
  2. Sodium is oxidised.

    Each sodium atom goes from 00 to +1+1, meaning it loses one electron: …

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